(i) Let \(y\) be the probability of throwing an odd number. Then the probability of throwing an even number is \(2y\). Since there are three odd and three even numbers, we have:
\(3y + 6y = 1\)
\(9y = 1\)
\(y = \frac{1}{9}\)
Thus, the probability of throwing an odd number is \(\frac{1}{3}\).
(ii) A score of 8 is obtained by throwing a 6. Since 6 is an even number, the probability \(P(X = 8) = \frac{2}{9}\).
(iii) The variance \(\text{Var}(X)\) is calculated as follows:
\(\text{Var}(X) = \frac{1}{9}(4^2 + 6^2 + 7^2 + 8^2 + 10^2) - \left(\frac{58}{9}\right)^2\)
\(= \frac{1}{9}(16 + 36 + 49 + 64 + 100) - \left(\frac{58}{9}\right)^2\)
\(= \frac{1}{9}(265) - \left(\frac{58}{9}\right)^2\)
\(= 29.44 - 41.78\)
\(= 4.02\)
(iv) The probability that the total of the scores on the two throws is 16 is:
\(P(6,10) + P(10,6) + P(8,8) = \frac{1}{81} + \frac{1}{81} + \frac{4}{81} = \frac{6}{81} = \frac{2}{27}\)
(v) Given that the total score is 16, the probability that the first score was 6 is:
\(P(6,10) = \frac{1}{81}\)
\(P(\text{first score 6 given total 16}) = \frac{\frac{1}{81}}{\frac{6}{81}} = \frac{1}{6}\)